Exercise 2.25

A finite number of neighborhoods of radius $\delta$ cover $K$. By the reasoning in ex. 24, $K$ has a countable base.

To see that this implies that $K$ is separable, let $\{U_n\}$ be a countable base, and choose $x_n\in U_n$, so that we obtain a countable subset $S={\{x_n\}}_{n=1}^{\infty }$. Any open set in $K$ contains $U_n$, and therefore an element of $S$. The $S$ is therefore dense in $K$.